Answer
Convergent
Work Step by Step
The Comparison Test states that the p-series $\sum_{n=1}^{\infty}\frac{1}{n^{p}}$ is convergent if $p\gt 1$ and divergent if $p\leq 1$.
$\Sigma _{k=1}^{\infty}\frac{(2k-1)(k^{2}-1)}{(k+1)(k^{2}+4)^{4}} \lt \Sigma _{k=1}^{\infty}\frac{(2k)(k^{2})}{(k+1)(k^{2}+4)^{4}}\lt \Sigma _{k=1}^{\infty}\frac{2k(k^{2})}{k(k^{2})^{2}} $
Note that
$\Sigma _{k=1}^{\infty}\frac{2k(k^{2})}{k(k^{2})^{2}}=\Sigma _{k=1}^{\infty}\frac{2k^{2}}{k^{5}} =2\Sigma _{k=1}^{\infty}\frac{1}{k^{2}} $
We find that the series is a p-series with $p=2 \gt 1$ , which is converging and we know that a series less than converging series is also converging.
